1

我有以下 JSON 对象，我想使用这些值来填充下拉列表。我一直试图让它工作 2 天，并准备把我的电脑扔出窗外。有人可以帮帮我吗？！

这是 JSON 对象...

{
"query":"de mar",
"suggestions":[
    "Any Location",
    "Camp De Mar, Majorca, SPAIN",
    "L'amettla De Mar, Costa Dorada, SPAIN",
    "Lloret De Mar, Costa Brava, SPAIN",
    "Malgrat De Mar, Costa Brava, SPAIN",
    "Pineda De Mar, Costa Brava, SPAIN",
    "Roquetas De Mar, Costa De Almeria, SPAIN",
    "Tossa De Mar, Costa Brava, SPAIN"
],
"data":[
    "",
    "DestinationResort|Camp De Mar",
    "DestinationResort|L'Amettla De Mar",
    "DestinationResort|Lloret De Mar",
    "DestinationResort|Malgrat De Mar",
    "DestinationResort|Pineda De Mar",
    "DestinationResort|Roquetas De Mar",
    "DestinationResort|Tossa De Mar"
]
}

还有我的 html 和 javascript...

<input type="text" id="search" data-provide="typeahead">

<select name="destinations" id="destinations"></select>

$("#search").on("input", function() {
$.getJSON( "http://holidays.allinclusive.co.uk/external/destinations.ashx?query=" + $("#search").val() )
.done(function(data) {
$.each(data, function() {
  $.each(this, function(name, value) {
    $( '<option value="' + data.value +'">' + value + '</option>' ).appendTo( "#destinations" );
  })
})
})
});

最终我正在寻找这样的东西......

<option value="DestinationResort|Camp De Mar">Camp De Mar, Majorca, SPAIN</option>

有人可以结束我的痛苦吗？

谢谢

4

3 回答 3

1

理想情况下，您可以重新格式化 JSON，以便将数据值和相应的文本组合在一起，而不是放在两个单独的数组中，但只要保证数组具有相同的索引，您就可以遍历一个并获取另一个的相应索引.

$.each(data.suggestions, function (idx, elem) {
    $("select").append(
        $("<option>").val(data.data[idx]).text(elem)
    );
});

http://jsfiddle.net/ExplosionPIlls/BwxtR/

于 2013-06-11T10:45:22.677 回答
0

这两个属性都指向数组，从我可以看到您的数组数据与索引匹配。所以这样的东西就足够了：

var option = '<option value="' + data.data[index] + '">" + data.suggestions[index] + "</option>";

获取您的代码：

$("#search").on("input", function() {
    $.getJSON( "http://holidays.allinclusive.co.uk/external/destinations.ashx?query=" + $("#search").val() )
        .done(function(data) {
            for (var index = 0; index < data.data.length; index++) {
                var option = '<option value="' + data.data[index] + '">" + data.suggestions[index] + "</option>";
            }
        })
    });
});
于 2013-06-11T10:44:14.563 回答
0

试试这个，

var json={"query":"de mar","suggestions":[    "Any Location",    "Camp De Mar, Majorca, SPAIN",    "L'amettla De Mar, Costa Dorada, SPAIN",    "Lloret De Mar, Costa Brava, SPAIN",    "Malgrat De Mar, Costa Brava, SPAIN",    "Pineda De Mar, Costa Brava, SPAIN",    "Roquetas De Mar, Costa De Almeria, SPAIN",    "Tossa De Mar, Costa Brava, SPAIN"],"data":[    "",    "DestinationResort|Camp De Mar",    "DestinationResort|L'Amettla De Mar",    "DestinationResort|Lloret De Mar",    "DestinationResort|Malgrat De Mar",    "DestinationResort|Pineda De Mar",    "DestinationResort|Roquetas De Mar",   "DestinationResort|Tossa De Mar"]};


$.each(json.suggestions, function(index,value) {
    $( '<option value="'+json.data[index]+'">'+value+'</option>').appendTo("#destinations" );
});

小提琴 http://jsfiddle.net/DbVUa/

于 2013-06-11T10:46:37.683 回答